Chapter 8.4, Problem 37E

Solution

a)

To find: The type of the conic section for the equation $2x^2+\sqrt{3}xy+y^2-10=0$ .

The graph of equation $2x^2+\sqrt{3}xy+y^2-10=0$ is an ellipse.

Given information:

The equation is $2x^2+\sqrt{3}xy+y^2-10=0$ .

Calculation:

Draw the graph of equation $2x^2+\sqrt{3}xy+y^2-10=0$

  

Therefore, the graph represents an ellipse.

b)

To find: The equation in terms of $\left(u,v\right)$ with no cross-product term and using axis of rotation.

The required equation in terms of $\left(u,v\right)$ is $\frac{{\text{v}}^{\text{2}}}{\text{20}}\text{+}\frac{{\text{u}}^{\text{2}}}{\text{4}}\text{=1}$ .

Given information:

The equation is $2x^2+\sqrt{3}xy+y^2-10=0$ .

Formula used:

Formula used to find the rotation angle $\text{θ}$ is:

  $\cot \theta =\frac{A-C}{B}$ , $0<\theta <\frac{\pi }{2}$

Here, A , B , and C are the values obtain by comparing given equation with equation $A{x}^2+Bxy+C{y}^2+Dx+Ey+F=0$ .

The equations for the axis of rotation are,

  $x=u\cos \theta -v\sin \theta$

  $y=u\sin \theta +v\cos \theta$

Calculation:

Comparing the given equation with the general equation,

  $A=2$

  $B=\sqrt{3}$

  $C=1$

Calculate the angle of rotation.

  $\begin{array}{c}\cot 2\theta =\frac{2-1}{\sqrt{3}}\\ =\frac{1}{\sqrt{3}}\end{array}$

Further, simplify.

  $\begin{array}{c}2\theta ={\cot }^{-1}\left(\frac{1}{\sqrt{3}}\right)\\ 2\theta =\frac{\pi }{3}\\ \theta =\frac{\pi }{6}\end{array}$

Substitute value of $\text{θ}$ in the equations of rotation of axis.

  $\begin{array}{c}x=u\cos \frac{\pi }{6}-v\sin \frac{\pi }{6}\\ x=\frac{\sqrt{3}}{2}u-\frac{v}{2}\end{array}$

And,

  $\begin{array}{c}y=u\sin \frac{\pi }{6}+v\cos \frac{\pi }{6}\\ y=\frac{u}{2}+\frac{\sqrt{3}}{2}v\end{array}$

Substituting the values of x and y in given equation.

  $\begin{array}{c}2{\left(\frac{\sqrt{3}}{2}u-\frac{v}{2}\right)}^2+\sqrt{3}\left(\frac{\sqrt{3}}{2}u-\frac{v}{2}\right)\left(\frac{u}{2}+\frac{\sqrt{3}}{2}v\right)+{\left(\frac{u}{2}+\frac{\sqrt{3}}{2}v\right)}^2-10=0\\ \frac{10}{4}{u}^2+\frac{2}{4}{v}^2=10\end{array}$

Dividing both sides by 10.

  $\frac{v^2}{20}+\frac{u^2}{4}=1$

The equation $2x^2+\sqrt{3}xy+y^2-10=0$ transfers in $\left(u,v\right)$ coordinates with no cross product as $\frac{v^2}{20}+\frac{u^2}{4}=1$ .

c)

To find: The vertex or vertices of equation $2x^2+\sqrt{3}xy+y^2-10=0$ in $\left(u,v\right)$ system.

The vertices of equation $2x^2+\sqrt{3}xy+y^2-10=0$ in $\left(u,v\right)$ system is: $\left(\text{0,2}\sqrt{\text{5}}\right)$ , $\left(\text{0,-2}\sqrt{\text{5}}\right)$

Given information:

The equation is $2x^2+\sqrt{3}xy+y^2-10=0$ .

The equation in $\left(u,v\right)$ system is,

  $\frac{v^2}{20}+\frac{u^2}{4}=1$

Formula used:

The standard formula for an ellipse is,

  $\frac{x^2}{b^2}+\frac{y^2}{a^2}=1$ , $a>b$

Vertices of ellipse are: $\left(\text{0,}\pm a\right)$

Calculation:

Comparing this equation with standard equation of ellipse.

  $\begin{array}{c}a=\sqrt{20}\\ =2\sqrt{5}\end{array}$

Therefore, the vertices of ellipse are $\left(\text{0,2}\sqrt{\text{5}}\right)$ and $\left(\text{0,-2}\sqrt{\text{5}}\right)$ .

d)

To find: The vertices of ellipse in $\left(x,y\right)$ system.

The vertices of equation $2x^2+\sqrt{3}xy+y^2-10=0$ are $\left(\text{-}\sqrt{\text{5}}\text{,}\sqrt{\text{15}}\right)$ and $\left(\sqrt{\text{5}}\text{,-}\sqrt{\text{15}}\right)$ .

Given information:

The equation is $2x^2+\sqrt{3}xy+y^2-10=0$ .

The equations for the rotation of axis are,

  $x=\frac{\sqrt{3}}{2}u-\frac{v}{2}$

  $y=\frac{u}{2}+\frac{\sqrt{3}}{2}v$

The vertices of equation $2x^2+\sqrt{3}xy+y^2-10=0$ in $\left(u,v\right)$ system are $\left(\text{0,2}\sqrt{\text{5}}\right)$ and $\left(\text{0,-2}\sqrt{\text{5}}\right)$ .

Calculation:

Write the equation of the rotation of axis in terms of x and y .

  $x=\frac{\sqrt{3}}{2}u-\frac{v}{2}$

  $y=\frac{u}{2}+\frac{\sqrt{3}}{2}v$

Substitute the values $u=0$ and $v=2\sqrt{5}$ ad v in the equation of rotation of axis.

  $\begin{array}{l}x=\frac{\sqrt{3}}{2}\left(0\right)-\frac{\left(2\sqrt{5}\right)}{2}\\ =\sqrt{5}\end{array}$

Similarly,

  $\begin{array}{c}y=\frac{\left(0\right)}{2}+\frac{\sqrt{3}}{2}\left(2\sqrt{5}\right)\\ =\sqrt{15}\end{array}$

Substitute the values $u=0$ and $v=-2\sqrt{5}$ ad v in the equation of rotation of axis.

  $\begin{array}{c}x=\frac{\sqrt{3}}{2}\left(0\right)-\frac{\left(-2\sqrt{5}\right)}{2}\\ =\sqrt{5}\end{array}$

Similarly,

  $\begin{array}{c}y=\frac{\left(0\right)}{2}+\frac{\sqrt{3}}{2}\left(-2\sqrt{5}\right)\\ =-\sqrt{15}\end{array}$

The vertices of equation $2x^2+\sqrt{3}xy+y^2-10=0$ are $\left(\text{-}\sqrt{\text{5}}\text{,}\sqrt{\text{15}}\right)$ and $\left(\sqrt{\text{5}}\text{,-}\sqrt{\text{15}}\right)$ .